7.2 Electromagnetics & Passive Components

7.2 Electromagnetics & Passive Components

Quick Answer: Faraday's Law ($e = -N \frac{d\Phi}{dt}$) and Lenz's Law define electromagnetic induction. Magnetic circuits operate via Ohm's Law analogue: Magnetomotive Force $\mathcal{F} = NI = \Phi \mathcal{R}$, where reluctance $\mathcal{R} = \frac{l}{\mu A}$. Passive components store energy in electric fields ($E_C = \frac{1}{2} C V^2$) and magnetic fields ($E_L = \frac{1}{2} L I^2$). First-order transient response is determined by time constants $\tau = RC$ (for RC circuits) and $\tau = \frac{L}{R}$ (for RL circuits), reaching steady state in $5\tau$. Second-order RLC transients are underdamped, critically damped, or overdamped based on damping factor $\alpha = \frac{R}{2L}$ vs natural frequency $\omega_0 = \frac{1}{\sqrt{LC}}$.

Fundamentals of Electromagnetics & Magnetic Circuits

Electromagnetism connects electric currents with magnetic fields, governing the behavior of inductors, transformers, relays, and electric motors.

1. Magnetic Field Quantities & Relations

• Magnetic Flux ($\Phi$): Total magnetic lines of force, measured in Webers (Wb).
• Magnetic Flux Density ($B$): Flux per unit area perpendicular to the magnetic field, measured in Tesla (T) or Wb/m$^2$: $B = \frac{\Phi}{A}$
• Magnetic Field Intensity ($H$): Magnetizing force per unit length, measured in A-turns/m: $H = \frac{N I}{l}$
• Permeability ($\mu$): Ability of a medium to conduct magnetic flux: $\mu = \mu_0 \mu_r$, where $\mu_0 = 4\pi \times 10^{-7}\text{ H/m}$ is free space permeability and $\mu_r$ is relative permeability. $B = \mu H = \mu_0 \mu_r H$

2. Electric vs. Magnetic Circuit Analogy

Magnetic circuits can be analyzed using an equivalent of Ohm's Law:

3. Faraday's & Lenz's Laws of Induction

• Faraday's Law: An electromotive force (EMF) is induced in a conductor or coil whenever the magnetic flux linking it changes over time: $e = -N \frac{d\Phi}{dt}$
• Lenz's Law: The minus sign represents Lenz's Law: the induced EMF generates a current whose magnetic field opposes the original change in magnetic flux that created it.
• Self-Inductance ($L$) & Mutual Inductance ($M$): $e_L = -L \frac{di}{dt} \implies L = \frac{N \Phi}{I} = \frac{N^2 \mu A}{l}$ $M = k \sqrt{L_1 L_2} \quad (0 \le k \le 1, \text{ magnetic coupling coefficient})$

---

Passive Components: Energy Storage & Properties

Resistors, capacitors, and inductors form the foundation of electronic circuits.

1. Capacitors & Energy Storage

A parallel-plate capacitor stores energy in an electric field between its dielectric-separated plates: $C = \frac{\epsilon A}{d} = \frac{\epsilon_0 \epsilon_r A}{d}$ where $\epsilon_0 = 8.854 \times 10^{-12}\text{ F/m}$.

• Instantaneous Current-Voltage Relation: $i(t) = C \frac{dv(t)}{dt}$
• Stored Energy ($E_C$): $E_C = \frac{1}{2} C V^2$ (Joules)
• Key Property: Voltage across a capacitor cannot change instantaneously ($v(0^+) = v(0^-)$).

2. Inductors & Energy Storage

An inductor stores energy in a magnetic field established by current flow through its turns:

• Instantaneous Current-Voltage Relation: $v(t) = L \frac{di(t)}{dt}$
• Stored Energy ($E_L$): $E_L = \frac{1}{2} L I^2$ (Joules)
• Key Property: Current through an inductor cannot change instantaneously ($i(0^+) = i(0^-)$).

---

First-Order & Second-Order Transient Responses

1. First-Order RC & RL Transient Response

When a step DC voltage $V_S$ is applied at $t=0$:

• Charging RC Circuit: $v_C(t) = V_S \left(1 - e^{-t/\tau}\right), \quad i_C(t) = \frac{V_S}{R} e^{-t/\tau} \quad (\text{where } \tau = RC)$
• Discharging RC Circuit: $v_C(t) = V_0 e^{-t/\tau}, \quad i_C(t) = -\frac{V_0}{R} e^{-t/\tau}$
• Energizing RL Circuit: $i_L(t) = \frac{V_S}{R} \left(1 - e^{-t/\tau}\right), \quad v_L(t) = V_S e^{-t/\tau} \quad \left(\text{where } \tau = \frac{L}{R}\right)$

2. Second-Order RLC Transient Response

The transient differential equation for a series RLC circuit driven by a step voltage is: $\frac{d^2 i}{dt^2} + \frac{R}{L}\frac{di}{dt} + \frac{1}{LC} i = 0$ The characteristic equation roots $s_{1,2}$ are: $s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2}$ where $\alpha = \frac{R}{2L}$ is the neper damping factor and $\omega_0 = \frac{1}{\sqrt{LC}}$ is the undamped resonant frequency.

Depending on the relationship between $\alpha$ and $\omega_0$, three transient response regimes exist:

1. Overdamped ($\alpha > \omega_0 \implies R > 2\sqrt{\frac{L}{C}}$): Real, distinct roots ($s_1, s_2$). Response returns to equilibrium slowly without oscillation.
2. Critically Damped ($\alpha = \omega_0 \implies R = 2\sqrt{\frac{L}{C}}$): Real, equal roots ($s_1 = s_2 = -\alpha$). Response returns to equilibrium in the shortest possible time without overshoot.
3. Underdamped ($\alpha < \omega_0 \implies R < 2\sqrt{\frac{L}{C}}$): Complex conjugate roots ($s_{1,2} = -\alpha \pm j\omega_d$). Response oscillates with damped natural frequency $\omega_d = \sqrt{\omega_0^2 - \alpha^2}$ while exponentially decaying.

---

Step-by-Step Worked Examples

Example 1: Magnetic Reluctance & Flux Calculation

Problem: An iron magnetic core has a mean path length $l = 0.5\text{ m}$, cross-sectional area $A = 10\text{ cm}^2$ ($10^{-3}\text{ m}^2$), relative permeability $\mu_r = 2000$, and a coil of $N = 500\text{ turns}$. Calculate: (a) Reluctance $\mathcal{R}$, and (b) Magnetic flux $\Phi$ when coil current $I = 2\text{ A}$.

Solution:

1. Calculate Reluctance $\mathcal{R}$: $\mathcal{R} = \frac{l}{\mu_0 \mu_r A} = \frac{0.5}{(4\pi \times 10^{-7}) \times (2000) \times (10^{-3})} = \frac{0.5}{8\pi \times 10^{-7}} = \frac{0.5}{2.5133 \times 10^{-6}} \approx 198,944\text{ A-t/Wb}$
2. Calculate MMF ($\mathcal{F}$): $\mathcal{F} = N I = 500 \times 2 = 1000\text{ A-turns}$
3. Calculate Flux ($\Phi$): $\Phi = \frac{\mathcal{F}}{\mathcal{R}} = \frac{1000}{198,944} \approx 5.027 \times 10^{-3}\text{ Wb} = 5.027\text{ mWb}$

Example 2: RC Transient Capacitor Voltage Calculation

Problem: A $100\ \mu\text{F}$ capacitor is charged through a $50\text{ k}\Omega$ resistor from a $12\text{ V}$ DC source. Determine: (a) Time constant $\tau$, and (b) Capacitor voltage at $t = 10\text{ seconds}$.

Solution:

1. Calculate Time Constant $\tau$: $\tau = R C = (50 \times 10^3\ \Omega) \times (100 \times 10^{-6}\text{ F}) = 5.0\text{ seconds}$
2. Calculate Voltage at $t = 10\text{ s}$ ($t = 2\tau$): $v_C(10) = 12 \left(1 - e^{-10/5}\right) = 12 \left(1 - e^{-2}\right) = 12 (1 - 0.1353) = 12 \times 0.8647 \approx 10.38\text{ V}$